高中数学题 急求解啊 坐等!帮忙!

1个回答

  • 1. 双曲线的渐近线: y = ±bx/a

    F(c, 0), P(√3/3, √6/3)在第一象限, 该渐近线是y = bx/a, 斜率k = b/a

    √6/3 = b(√3/3)/a

    b/a = √2, b = √2a

    PF的斜率-1/√2 = (√6/3- 0)/(c - √3/3)

    c = √3

    c² = a² + b² = a² + 2a² = 3a² = 3

    a = 1, b = √2

    双曲线方程: x² - y²/2 = 1

    2.

    设A(p, q), B(u, v)

    (p + u)/2 = 1, p + u = 2 (i)

    (q + v)/2 = 2, q + v = 4 (ii)

    p² - q²/2 = 1 (iii)

    u² - v²/2 = 1 (iv)

    (iii) - (iv): (p + u)(p -u) - (q + v)(q - v)/2 = 0

    2(p - u) - 4(q - v)/2 = 0

    (q - v)/(p - u) = 1

    此为AB的斜率, AB的方程: y - 2 = 1*(x - 1)

    y = x + 1